TITLE: How does multiplication affect degrees?
QUESTION [5 upvotes]: Let $M(n) \sim \mathbb{A}^{n^2}$ be the space of $n$-by-$n$ matrices, seen as an affine space over a field $K$, and endowed with the usual matrix multiplication. Let $V$ and $W$ be subvarieties of $M(n)$. The product $V\cdot W$ is a constructible set (by Chevalley's theorem); write $\overline{V\cdot W}$ for its Zariski closure.
Is it the case that $\deg(\overline{V\cdot W}) \leq \deg(V) \cdot \deg(W)$?

Note: this is a less ambitious version of a question I previously asked (Bézout and products in algebraic groups).

REPLY [5 votes]: This is false.  Let $G$ be the open subset $\textbf{GL}_2$ in the affine space $\textbf{Mat}_{2\times 2} = \mathbb{A}^4$ with affine coordinates $x_{1,1}$, $x_{2,1}$, $x_{1,2}$ and $x_{2,2}$.  Let $V$ be the one-dimensional subvariety,
$$
V = \text{Zero}(x_{2,1},x_{2,2}-1,x_{1,1}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & s \\ 0 & 1 \end{array} \right] : s\in \mathbb{A}^1 \right\}.
$$
Similarly, let $W$ be the following one-dimensional subvariety,
$$
V = \text{Zero}(x_{2,1},x_{1,1}-1,x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} 1 & t \\ 0 & t \end{array} \right] : t\in \mathbb{A}^1 \right\}.
$$
Each of $V$ and $W$ is a one-dimensional, affine linear variety.  In particular, each has degree $1$.  Yet the product set $\overline{V\cdot W}$ is the following two-dimensional variety,
$$
W = \text{Zero}(x_{2,1},2x_{1,1}x_{2,2}-x_{1,2}) = \left\{\left[ \begin{array}{rr} s & 2st \\ 0 & t \end{array} \right] : (s,t)\in \mathbb{A}^2 \right\}.
$$
This has degree $2$.